Question

What is t squared minus 36 over 3 divided by t squared plus 6t over 9t

Answer 1

Answer: 3t-18

Explanation:

`((t^2-36)/(3))/((t^2+6t)/(9t) )`

The first step is to multiply by the reciprocal of the denominator. The reciprocal of a fraction is its inverted fraction which will make it equal to 1.

`((t^2-36)/(3))/((t^2+6t)/(9t) )*(9t)/(t^2+6t)`

This will eliminate the denominator,leaving the fraction like;

`(t^2-36)/(3)*(9t)/(t^2+6t)`

You can simplify 9 and 3;

9/3=3

3/3=1

`t^2-36*(3t)/(t^2+6t)`

Multiply;

`(3t^3-108t)/(t^2+6t)`

We can factor both numerator and denominator. The common factor in the numerator is 3t. And, the common factor in the denominator is t.

`(3t(t^2-36))/(t(t+6))`

Now we can simplify t and t.

`(3(t^2-36))/(t+6)`

`t^2` and
`36` are both perfect squares with a negative (-) sign in the middle, in this case, we can extract both squares and multiply them by themselves being one positive and the other negative.

`(3[(t+6)(t-6)])/(t+6)`

Since t + 6 and t - 7 are being multiplied, it is possible to simplify with the denominator. Leaving the fraction like;

`3(t-6)`

Multiply;

`3t-18`